The line $y = b-x$ with $0 < b < 4$ intersects the $y$-axis at $P$ and the line $x=4$ at $S$. If the ratio of the area of triangle $QRS$ to the area of triangle $QOP$ is 9:25, what is the value of $b$? Express the answer as a decimal to the nearest tenth.

Answer:b=2.5Step-by-step explanation:The given line has equation: [tex]y=b-x[/tex]The x-intercept is (b,0)The y-intercept is (0,b)The area of triangle QOP [tex]=\frac{1}{2} |OQ|*|OP|=[/tex] [tex]=\frac{1}{2}b*b[/tex] [tex]=\frac{1}{2}b^2[/tex]The area of triangle QRS [tex]=\frac{1}{2} |RQ|*|RS|=[/tex] [tex]=\frac{1}{2}(4-b)*(b-4)[/tex]The area of triangle QRS and QOP are in the ratio 9:25We must negate the area under the x-axis.[tex]\implies \frac{- \frac{1}{2}(4-b)(b-4)}{\frac{1}{2}b^2} =\frac{9}{25}[/tex][tex]\implies \frac{-(4-b)(b-4)}{b^2} =\frac{9}{25}[/tex][tex]\implies {-(4-b)(b-4) =\frac{9}{25}b^2[/tex][tex]\implies b^2-8b+16=\frac{9}{25}b^2[/tex][tex]\implies 25b^2-200b+400=9b^2[/tex][tex]\implies 25b^2-9b^2-200b+400=0[/tex][tex]\implies 16b^2-200b+400=0[/tex][tex]\implies 2b^2-25b+50=0[/tex][tex](2b-5)(b-10)=0[/tex][tex]b=2.5\:or\:b=10[/tex]But 0<b<4Therefore b=2.5VERIFYThe area of triangle QOP [tex]=\frac{1}{2}*2.5*2.5=3.125[/tex]The area of triangle QRS [tex]=|\frac{1}{2}(4-2.5)*(2.5-4)|=|-1.125|=1.125[/tex]We must take absolute value because this area is below the x-axis.[tex]\frac{1.125}{3.125} =\frac{9}{25}[/tex]